Related
The problem definition is that
Given an additional digit 0 ≤ x ≤ 9, write a function that returns
the integer that results from inserting x in n, such that its digits
also appear in ascending order from left to right. For instance, if n
= 24667 and x = 5, the function should return 245667.
My code
// the divisions are integer division, no floating point
int x(int n, int insertValue)
{
if (n == 0) return 0;
int val = x(n/10, insertValue);
if((n%10) > insertValue)
{
int q = insertValue * 10 + (n%10);
return val * 100 + q;
}
return val*10 + (n%10);
}
For the case of, for example, x(2245,3), it outputs 223435. But I have already done with it while processing 224. It shouldn't go on adding the value to be inserted any more, I mean the 3 shouldn't be there before 5 .
I can come up with a solution that I can put in each recursion step a boolean flag that identify by taking modulo by 10 and dividing by 10 up to reach the single digit case. If there is no any identification, go in to that if block, else not. But it sounds too silly.
When you're dividing recursively, you are actually going right to left, not left to right, so you should check if a digit is smaller than the one inserted and not greater (unless you always let the recursion reach n==0 condition and make your comparisons on your way out of it but that would be ineffective).
The second thing is that you do not break recursion once you inserted the digit (which you were aware of as I can see now in the question title), so it gets inserted repeatedly before every digit that is larger than insertValue. As to how to do it: you were already stopping recursion with if(n==0) condition, i.e. if n==0 the function stops calling itself (it returns immediately). When inserting the digit the difference is that you need to use the original value (n) to return from the function instead of passing it further.
At this point it works well for your example but there's also one edge case you need to consider if you want the function to work property. When you have nothing to divide anymore [if(n==0)] you need to insert your digit anyway (return insertValue) so it does not get lost on the left edge like in x(2245,1) call.
Improvements for brevity:
% has the same precedence as * and /, so brackets around it are not needed here.
I removed val variable as it was now used only once and its calculation was not always necessary.
Here's the working code:
int x(int n, int insertValue){
if(n == 0) return insertValue;
//if insertion point was found, use original value (n)
if(n%10 <= insertValue)
return n*10 + insertValue;
//if not there yet, keep calling x()
return x(n/10, insertValue)*10 + n%10;
}
I'm having trouble understanding how the recursive merge sort algorithm works, I understand how it theoretically works :
if there's more than one element in an array find its middle and divide the array into 2 smaller sub-arrays and so on until you have 2 arrays of 1 element that are by definition already sorted (base case) then you can merge them using a merging algorithm, then you go up the tree and so on.
I've tried to implement it in python with some print statements to follow step by step and it works but I don't really understand why it works the way it does. I'll describe you my wrong logic:
the algorithm in pseudo code with l being the low index and h the high index :
merge_sort(l,h){
if(l < h){
mid = (l+h)//2
merge_sort(l,mid)
merge_sort(mid+1,h)
merge(l,mid,h)
}
}
so for a an arr = [9,3,7,5]
We call merge_sort(0,3) which is the full array, l=0 , h =3, mid = 1 [9,3,7,5]
l < h so it calls merge_sort(0,1), l=0,h=1 , mid=0 [9,3]
l < h so it calls merge_sort(0,0) l=0 h=0 mid =0 [9]
from here l == h so we don't fulfill the condition l < h and thus merge_sort doesn't call itself again, for me the algorithm ends here ( which is obviously wrong ) BUT here it goes upand calls merge_sort(1,1) [3] (right side of the array) then merge is executed.
After that it goes up again and tackles the right side of the initial array merge_sort(2,3) [7,5] and continues. ( [7] then [5] then merging )
Can someone please explain me why the algorithms continues to call himself after l == n ? I've read some explanations and watch some videos and some even argues merge(l,mid) sort the full first half [9,3] and merge(mid+1,h) sort the second full half [7,5] which is apparently not how it's working, I'm really confused.
It doesn't call itself again, it returns back to it's previous step in recursion, once it finishes the current one.
Here is an example from GeeksForGeeks of a merge sort algorithm, with each step labeled numerically in a sequential order:
You can understand a recursive algorithm by trusting it. Have a look at the annotation below (! starts a comment):
merge_sort(l,h) {
if (l < h) {
mid = (l+h)//2
merge_sort(l,mid) ! If merge_sort does its job, the array is now sorted from l to mid
merge_sort(mid+1,h) ! If merge_sort does its job, the array is now sorted from mid+1 to h
merge(l,mid,h) ! If merge does its job, the array is now sorted from l to h
}
else { ! If l==h, the array is already sorted from l to h
}
Now you can see that if we call the function with some values of l and h, it will return a sorted array, by calling itself on parts of the array, each time returning them sorted. The whole construction works because the size of the subarrays goes decreasing across the levels of calls, and always reaches 1, causing an immediate return to the calling level.
I'm working on a homework problem that asks me this:
Tiven a finite set of numbers, and a target number, find if the set can be used to calculate the target number using basic math operations (add, sub, mult, div) and using each number in the set exactly once (so I need to exhaust the set). This has to be done with recursion.
So, for example, if I have the set
{1, 2, 3, 4}
and target 10, then I could get to it by using
((3 * 4) - 2)/1 = 10.
I'm trying to phrase the algorithm in pseudo-code, but so far haven't gotten too far. I'm thinking graphs are the way to go, but would definitely appreciate help on this. thanks.
This isn't meant to be the fastest solution, but rather an instructive one.
It recursively generates all equations in postfix notation
It also provides a translation from postfix to infix notation
There is no actual arithmetic calculation done, so you have to implement that on your own
Be careful about division by zero
With 4 operands, 4 possible operators, it generates all 7680 = 5 * 4! * 4^3
possible expressions.
5 is Catalan(3). Catalan(N) is the number of ways to paranthesize N+1 operands.
4! because the 4 operands are permutable
4^3 because the 3 operators each have 4 choice
This definitely does not scale well, as the number of expressions for N operands is [1, 8, 192, 7680, 430080, 30965760, 2724986880, ...].
In general, if you have n+1 operands, and must insert n operators chosen from k possibilities, then there are (2n)!/n! k^n possible equations.
Good luck!
import java.util.*;
public class Expressions {
static String operators = "+-/*";
static String translate(String postfix) {
Stack<String> expr = new Stack<String>();
Scanner sc = new Scanner(postfix);
while (sc.hasNext()) {
String t = sc.next();
if (operators.indexOf(t) == -1) {
expr.push(t);
} else {
expr.push("(" + expr.pop() + t + expr.pop() + ")");
}
}
return expr.pop();
}
static void brute(Integer[] numbers, int stackHeight, String eq) {
if (stackHeight >= 2) {
for (char op : operators.toCharArray()) {
brute(numbers, stackHeight - 1, eq + " " + op);
}
}
boolean allUsedUp = true;
for (int i = 0; i < numbers.length; i++) {
if (numbers[i] != null) {
allUsedUp = false;
Integer n = numbers[i];
numbers[i] = null;
brute(numbers, stackHeight + 1, eq + " " + n);
numbers[i] = n;
}
}
if (allUsedUp && stackHeight == 1) {
System.out.println(eq + " === " + translate(eq));
}
}
static void expression(Integer... numbers) {
brute(numbers, 0, "");
}
public static void main(String args[]) {
expression(1, 2, 3, 4);
}
}
Before thinking about how to solve the problem (like with graphs), it really helps to just look at the problem. If you find yourself stuck and can't seem to come up with any pseudo-code, then most likely there is something that is holding you back; Some other question or concern that hasn't been addressed yet. An example 'sticky' question in this case might be, "What exactly is recursive about this problem?"
Before you read the next paragraph, try to answer this question first. If you knew what was recursive about the problem, then writing a recursive method to solve it might not be very difficult.
You want to know if some expression that uses a set of numbers (each number used only once) gives you a target value. There are four binary operations, each with an inverse. So, in other words, you want to know if the first number operated with some expression of the other numbers gives you the target. Well, in other words, you want to know if some expression of the 'other' numbers is [...]. If not, then using the first operation with the first number doesn't really give you what you need, so try the other ops. If they don't work, then maybe it just wasn't meant to be.
Edit: I thought of this for an infix expression of four operators without parenthesis, since a comment on the original question said that parenthesis were added for the sake of an example (for clarity?) and the use of parenthesis was not explicitly stated.
Well, you didn't mention efficiency so I'm going to post a really brute force solution and let you optimize it if you want to. Since you can have parantheses, it's easy to brute force it using Reverse Polish Notation:
First of all, if your set has n numbers, you must use exactly n - 1 operators. So your solution will be given by a sequence of 2n - 1 symbols from {{your given set}, {*, /, +, -}}
st = a stack of length 2n - 1
n = numbers in your set
a = your set, to which you add *, /, +, -
v[i] = 1 if the NUMBER i has been used before, 0 otherwise
void go(int k)
{
if ( k > 2n - 1 )
{
// eval st as described on Wikipedia.
// Careful though, it might not be valid, so you'll have to check that it is
// if it evals to your target value great, you can build your target from the given numbers. Otherwise, go on.
return;
}
for ( each symbol x in a )
if ( x isn't a number or x is a number but v[x] isn't 1 )
{
st[k] = x;
if ( x is a number )
v[x] = 1;
go(k + 1);
}
}
Generally speaking, when you need to do something recursively it helps to start from the "bottom" and think your way up.
Consider: You have a set S of n numbers {a,b,c,...}, and a set of four operations {+,-,*,/}. Let's call your recursive function that operates on the set F(S)
If n is 1, then F(S) will just be that number.
If n is 2, F(S) can be eight things:
pick your left-hand number from S (2 choices)
then pick an operation to apply (4 choices)
your right-hand number will be whatever is left in the set
Now, you can generalize from the n=2 case:
Pick a number x from S to be the left-hand operand (n choices)
Pick an operation to apply
your right hand number will be F(S-x)
I'll let you take it from here. :)
edit: Mark poses a valid criticism; the above method won't get absolutely everything. To fix that problem, you need to think about it in a slightly different way:
At each step, you first pick an operation (4 choices), and then
partition S into two sets, for the left and right hand operands,
and recursively apply F to both partitions
Finding all partitions of a set into 2 parts isn't trivial itself, though.
Your best clue about how to approach this problem is the fact that your teacher/professor wants you to use recursion. That is, this isn't a math problem - it is a search problem.
Not to give too much away (it is homework after all), but you have to spawn a call to the recursive function using an operator, a number and a list containing the remaining numbers. The recursive function will extract a number from the list and, using the operation passed in, combine it with the number passed in (which is your running total). Take the running total and call yourself again with the remaining items on the list (you'll have to iterate the list within the call but the sequence of calls is depth-first). Do this once for each of the four operators unless Success has been achieved by a previous leg of the search.
I updated this to use a list instead of a stack
When the result of the operation is your target number and your list is empty, then you have successfully found the set of operations (those that traced the path to the successful leaf) - set the Success flag and unwind. Note that the operators aren't on a list nor are they in the call: the function itself always iterates over all four. Your mechanism for "unwinding" the operator sequence from the successful leaf to get the sequence is to return the current operator and number prepended to the value returned by recursive call (only one of which will be successful since you stop at success - that, obviously, is the one to use). If none are successful, then what you return isn't important anyhow.
Update This is much harder when you have to consider expressions like the one that Daniel posted. You have combinatorics on the numbers and the groupings (numbers due to the fact that / and - are order sensitive even without grouping and grouping because it changes precedence). Then, of course, you also have the combinatorics of the operations. It is harder to manage the differences between (4 + 3) * 2 and 4 + (3 * 2) because grouping doesn't recurse like operators or numbers (which you can just iterate over in a breadth-first manner while making your (depth-first) recursive calls).
Here's some Python code to get you started: it just prints all the possible expressions, without worrying too much about redundancy. You'd need to modify it to evaluate expressions and compare to the target number, rather than simply printing them.
The basic idea is: given a set S of numbers, partition S into two subsets left and right in all possible ways (where we don't care about the order or the elements in left and right), such that left and right are both nonempty. Now for each of these partitions, find all ways of combining the elements in left (recursively!), and similarly for right, and combine the two resulting values with all possible operators. The recursion bottoms out when a set has just one element, in which case there's only one value possible.
Even if you don't know Python, the expressions function should be reasonably easy to follow; the splittings function contains some Python oddities, but all it does is to find all the partitions of the list l into left and right pieces.
def splittings(l):
n = len(l)
for i in xrange(2**n):
left = [e for b, e in enumerate(l) if i & 2**b]
right = [e for b, e in enumerate(l) if not i & 2**b]
yield left, right
def expressions(l):
if len(l) == 1:
yield l[0]
else:
for left, right in splittings(l):
if not left or not right:
continue
for el in expressions(left):
for er in expressions(right):
for operator in '+-*/':
yield '(' + el + operator + er + ')'
for x in expressions('1234'):
print x
pusedo code:
Works(list, target)
for n in list
tmp=list.remove(n)
return Works(tmp,target+n) or Works(tmp,target-n) or Works(tmp, n-target) or ...
then you just have to put the base case in. I think I gave away to much.
I'm working on a homework problem that asks me this:
Tiven a finite set of numbers, and a target number, find if the set can be used to calculate the target number using basic math operations (add, sub, mult, div) and using each number in the set exactly once (so I need to exhaust the set). This has to be done with recursion.
So, for example, if I have the set
{1, 2, 3, 4}
and target 10, then I could get to it by using
((3 * 4) - 2)/1 = 10.
I'm trying to phrase the algorithm in pseudo-code, but so far haven't gotten too far. I'm thinking graphs are the way to go, but would definitely appreciate help on this. thanks.
This isn't meant to be the fastest solution, but rather an instructive one.
It recursively generates all equations in postfix notation
It also provides a translation from postfix to infix notation
There is no actual arithmetic calculation done, so you have to implement that on your own
Be careful about division by zero
With 4 operands, 4 possible operators, it generates all 7680 = 5 * 4! * 4^3
possible expressions.
5 is Catalan(3). Catalan(N) is the number of ways to paranthesize N+1 operands.
4! because the 4 operands are permutable
4^3 because the 3 operators each have 4 choice
This definitely does not scale well, as the number of expressions for N operands is [1, 8, 192, 7680, 430080, 30965760, 2724986880, ...].
In general, if you have n+1 operands, and must insert n operators chosen from k possibilities, then there are (2n)!/n! k^n possible equations.
Good luck!
import java.util.*;
public class Expressions {
static String operators = "+-/*";
static String translate(String postfix) {
Stack<String> expr = new Stack<String>();
Scanner sc = new Scanner(postfix);
while (sc.hasNext()) {
String t = sc.next();
if (operators.indexOf(t) == -1) {
expr.push(t);
} else {
expr.push("(" + expr.pop() + t + expr.pop() + ")");
}
}
return expr.pop();
}
static void brute(Integer[] numbers, int stackHeight, String eq) {
if (stackHeight >= 2) {
for (char op : operators.toCharArray()) {
brute(numbers, stackHeight - 1, eq + " " + op);
}
}
boolean allUsedUp = true;
for (int i = 0; i < numbers.length; i++) {
if (numbers[i] != null) {
allUsedUp = false;
Integer n = numbers[i];
numbers[i] = null;
brute(numbers, stackHeight + 1, eq + " " + n);
numbers[i] = n;
}
}
if (allUsedUp && stackHeight == 1) {
System.out.println(eq + " === " + translate(eq));
}
}
static void expression(Integer... numbers) {
brute(numbers, 0, "");
}
public static void main(String args[]) {
expression(1, 2, 3, 4);
}
}
Before thinking about how to solve the problem (like with graphs), it really helps to just look at the problem. If you find yourself stuck and can't seem to come up with any pseudo-code, then most likely there is something that is holding you back; Some other question or concern that hasn't been addressed yet. An example 'sticky' question in this case might be, "What exactly is recursive about this problem?"
Before you read the next paragraph, try to answer this question first. If you knew what was recursive about the problem, then writing a recursive method to solve it might not be very difficult.
You want to know if some expression that uses a set of numbers (each number used only once) gives you a target value. There are four binary operations, each with an inverse. So, in other words, you want to know if the first number operated with some expression of the other numbers gives you the target. Well, in other words, you want to know if some expression of the 'other' numbers is [...]. If not, then using the first operation with the first number doesn't really give you what you need, so try the other ops. If they don't work, then maybe it just wasn't meant to be.
Edit: I thought of this for an infix expression of four operators without parenthesis, since a comment on the original question said that parenthesis were added for the sake of an example (for clarity?) and the use of parenthesis was not explicitly stated.
Well, you didn't mention efficiency so I'm going to post a really brute force solution and let you optimize it if you want to. Since you can have parantheses, it's easy to brute force it using Reverse Polish Notation:
First of all, if your set has n numbers, you must use exactly n - 1 operators. So your solution will be given by a sequence of 2n - 1 symbols from {{your given set}, {*, /, +, -}}
st = a stack of length 2n - 1
n = numbers in your set
a = your set, to which you add *, /, +, -
v[i] = 1 if the NUMBER i has been used before, 0 otherwise
void go(int k)
{
if ( k > 2n - 1 )
{
// eval st as described on Wikipedia.
// Careful though, it might not be valid, so you'll have to check that it is
// if it evals to your target value great, you can build your target from the given numbers. Otherwise, go on.
return;
}
for ( each symbol x in a )
if ( x isn't a number or x is a number but v[x] isn't 1 )
{
st[k] = x;
if ( x is a number )
v[x] = 1;
go(k + 1);
}
}
Generally speaking, when you need to do something recursively it helps to start from the "bottom" and think your way up.
Consider: You have a set S of n numbers {a,b,c,...}, and a set of four operations {+,-,*,/}. Let's call your recursive function that operates on the set F(S)
If n is 1, then F(S) will just be that number.
If n is 2, F(S) can be eight things:
pick your left-hand number from S (2 choices)
then pick an operation to apply (4 choices)
your right-hand number will be whatever is left in the set
Now, you can generalize from the n=2 case:
Pick a number x from S to be the left-hand operand (n choices)
Pick an operation to apply
your right hand number will be F(S-x)
I'll let you take it from here. :)
edit: Mark poses a valid criticism; the above method won't get absolutely everything. To fix that problem, you need to think about it in a slightly different way:
At each step, you first pick an operation (4 choices), and then
partition S into two sets, for the left and right hand operands,
and recursively apply F to both partitions
Finding all partitions of a set into 2 parts isn't trivial itself, though.
Your best clue about how to approach this problem is the fact that your teacher/professor wants you to use recursion. That is, this isn't a math problem - it is a search problem.
Not to give too much away (it is homework after all), but you have to spawn a call to the recursive function using an operator, a number and a list containing the remaining numbers. The recursive function will extract a number from the list and, using the operation passed in, combine it with the number passed in (which is your running total). Take the running total and call yourself again with the remaining items on the list (you'll have to iterate the list within the call but the sequence of calls is depth-first). Do this once for each of the four operators unless Success has been achieved by a previous leg of the search.
I updated this to use a list instead of a stack
When the result of the operation is your target number and your list is empty, then you have successfully found the set of operations (those that traced the path to the successful leaf) - set the Success flag and unwind. Note that the operators aren't on a list nor are they in the call: the function itself always iterates over all four. Your mechanism for "unwinding" the operator sequence from the successful leaf to get the sequence is to return the current operator and number prepended to the value returned by recursive call (only one of which will be successful since you stop at success - that, obviously, is the one to use). If none are successful, then what you return isn't important anyhow.
Update This is much harder when you have to consider expressions like the one that Daniel posted. You have combinatorics on the numbers and the groupings (numbers due to the fact that / and - are order sensitive even without grouping and grouping because it changes precedence). Then, of course, you also have the combinatorics of the operations. It is harder to manage the differences between (4 + 3) * 2 and 4 + (3 * 2) because grouping doesn't recurse like operators or numbers (which you can just iterate over in a breadth-first manner while making your (depth-first) recursive calls).
Here's some Python code to get you started: it just prints all the possible expressions, without worrying too much about redundancy. You'd need to modify it to evaluate expressions and compare to the target number, rather than simply printing them.
The basic idea is: given a set S of numbers, partition S into two subsets left and right in all possible ways (where we don't care about the order or the elements in left and right), such that left and right are both nonempty. Now for each of these partitions, find all ways of combining the elements in left (recursively!), and similarly for right, and combine the two resulting values with all possible operators. The recursion bottoms out when a set has just one element, in which case there's only one value possible.
Even if you don't know Python, the expressions function should be reasonably easy to follow; the splittings function contains some Python oddities, but all it does is to find all the partitions of the list l into left and right pieces.
def splittings(l):
n = len(l)
for i in xrange(2**n):
left = [e for b, e in enumerate(l) if i & 2**b]
right = [e for b, e in enumerate(l) if not i & 2**b]
yield left, right
def expressions(l):
if len(l) == 1:
yield l[0]
else:
for left, right in splittings(l):
if not left or not right:
continue
for el in expressions(left):
for er in expressions(right):
for operator in '+-*/':
yield '(' + el + operator + er + ')'
for x in expressions('1234'):
print x
pusedo code:
Works(list, target)
for n in list
tmp=list.remove(n)
return Works(tmp,target+n) or Works(tmp,target-n) or Works(tmp, n-target) or ...
then you just have to put the base case in. I think I gave away to much.
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I'm having major trouble understanding recursion at school. Whenever the professor is talking about it, I seem to get it but as soon as I try it on my own it completely blows my brains.
I was trying to solve Towers of Hanoi all night and completely blew my mind. My textbook has only about 30 pages in recursion so it is not too useful. Does anyone know of books or resources that can help clarify this topic?
How do you empty a vase containing five flowers?
Answer: if the vase is not empty, you take out one flower
and then you empty a vase containing four flowers.
How do you empty a vase containing four flowers?
Answer: if the vase is not empty, you take out one flower
and then you empty a vase containing three flowers.
How do you empty a vase containing three flowers?
Answer: if the vase is not empty, you take out one flower
and then you empty a vase containing two flowers.
How do you empty a vase containing two flowers?
Answer: if the vase is not empty, you take out one flower
and then you empty a vase containing one flower.
How do you empty a vase containing one flower?
Answer: if the vase is not empty, you take out one flower
and then you empty a vase containing no flowers.
How do you empty a vase containing no flowers?
Answer: if the vase is not empty, you take out one flower
but the vase is empty so you're done.
That's repetitive. Let's generalize it:
How do you empty a vase containing N flowers?
Answer: if the vase is not empty, you take out one flower
and then you empty a vase containing N-1 flowers.
Hmm, can we see that in code?
void emptyVase( int flowersInVase ) {
if( flowersInVase > 0 ) {
// take one flower and
emptyVase( flowersInVase - 1 ) ;
} else {
// the vase is empty, nothing to do
}
}
Hmm, couldn't we have just done that in a for loop?
Why, yes, recursion can be replaced with iteration, but often recursion is more elegant.
Let's talk about trees. In computer science, a tree is a structure made up of nodes, where each node has some number of children that are also nodes, or null. A binary tree is a tree made of nodes that have exactly two children, typically called "left" and "right"; again the children can be nodes, or null. A root is a node that is not the child of any other node.
Imagine that a node, in addition to its children, has a value, a number, and imagine that we wish to sum all the values in some tree.
To sum value in any one node, we would add the value of node itself to the value of its left child, if any, and the value of its right child, if any. Now recall that the children, if they're not null, are also nodes.
So to sum the left child, we would add the value of child node itself to the value of its left child, if any, and the value of its right child, if any.
So to sum the value of the left child's left child, we would add the value of child node itself to the value of its left child, if any, and the value of its right child, if any.
Perhaps you've anticipated where I'm going with this, and would like to see some code? OK:
struct node {
node* left;
node* right;
int value;
} ;
int sumNode( node* root ) {
// if there is no tree, its sum is zero
if( root == null ) {
return 0 ;
} else { // there is a tree
return root->value + sumNode( root->left ) + sumNode( root->right ) ;
}
}
Notice that instead of explicitly testing the children to see if they're null or nodes, we just make the recursive function return zero for a null node.
So say we have a tree that looks like this (the numbers are values, the slashes point to children, and # means the pointer points to null):
5
/ \
4 3
/\ /\
2 1 # #
/\ /\
## ##
If we call sumNode on the root (the node with value 5), we will return:
return root->value + sumNode( root->left ) + sumNode( root->right ) ;
return 5 + sumNode( node-with-value-4 ) + sumNode( node-with-value-3 ) ;
Let's expand that in place. Everywhere we see sumNode, we'll replace it with the expansion of the return statement:
sumNode( node-with-value-5);
return root->value + sumNode( root->left ) + sumNode( root->right ) ;
return 5 + sumNode( node-with-value-4 ) + sumNode( node-with-value-3 ) ;
return 5 + 4 + sumNode( node-with-value-2 ) + sumNode( node-with-value-1 )
+ sumNode( node-with-value-3 ) ;
return 5 + 4
+ 2 + sumNode(null ) + sumNode( null )
+ sumNode( node-with-value-1 )
+ sumNode( node-with-value-3 ) ;
return 5 + 4
+ 2 + 0 + 0
+ sumNode( node-with-value-1 )
+ sumNode( node-with-value-3 ) ;
return 5 + 4
+ 2 + 0 + 0
+ 1 + sumNode(null ) + sumNode( null )
+ sumNode( node-with-value-3 ) ;
return 5 + 4
+ 2 + 0 + 0
+ 1 + 0 + 0
+ sumNode( node-with-value-3 ) ;
return 5 + 4
+ 2 + 0 + 0
+ 1 + 0 + 0
+ 3 + sumNode(null ) + sumNode( null ) ;
return 5 + 4
+ 2 + 0 + 0
+ 1 + 0 + 0
+ 3 + 0 + 0 ;
return 5 + 4
+ 2 + 0 + 0
+ 1 + 0 + 0
+ 3 ;
return 5 + 4
+ 2 + 0 + 0
+ 1
+ 3 ;
return 5 + 4
+ 2
+ 1
+ 3 ;
return 5 + 4
+ 3
+ 3 ;
return 5 + 7
+ 3 ;
return 5 + 10 ;
return 15 ;
Now see how we conquered a structure of arbitrary depth and "branchiness", by considering it as the repeated application of a composite template? each time through our sumNode function, we dealt with only a single node, using a single if/then branch, and two simple return statements that almost wrote themsleves, directly from our specification?
How to sum a node:
If a node is null
its sum is zero
otherwise
its sum is its value
plus the sum of its left child node
plus the sum of its right child node
That's the power of recursion.
The vase example above is an example of tail recursion. All that tail recursion means is that in the recursive function, if we recursed (that is, if we called the function again), that was the last thing we did.
The tree example was not tail recursive, because even though that last thing we did was to recurse the right child, before we did that we recursed the left child.
In fact, the order in which we called the children, and added the current node's value didn't matter at all, because addition is commutative.
Now let's look at an operation where order does matter. We'll use a binary tree of nodes, but this time the value held will be a character, not a number.
Our tree will have a special property, that for any node, its character comes after (in alphabetical order) the character held by its left child and before (in alphabetical order) the character held by its right child.
What we want to do is print the tree in alphabetical order. That's easy to do, given the tree special property. We just print the left child, then the node's character, then right child.
We don't just want to print willy-nilly, so we'll pass our function something to print on. This will be an object with a print( char ) function; we don't need to worry about how it works, just that when print is called, it'll print something, somewhere.
Let's see that in code:
struct node {
node* left;
node* right;
char value;
} ;
// don't worry about this code
class Printer {
private ostream& out;
Printer( ostream& o ) :out(o) {}
void print( char c ) { out << c; }
}
// worry about this code
int printNode( node* root, Printer& printer ) {
// if there is no tree, do nothing
if( root == null ) {
return ;
} else { // there is a tree
printNode( root->left, printer );
printer.print( value );
printNode( root->right, printer );
}
Printer printer( std::cout ) ;
node* root = makeTree() ; // this function returns a tree, somehow
printNode( root, printer );
In addition to the order of operations now mattering, this example illustrates that we can pass things into a recursive function. The only thing we have to do is make sure that on each recursive call, we continue to pass it along. We passed in a node pointer and a printer to the function, and on each recursive call, we passed them "down".
Now if our tree looks like this:
k
/ \
h n
/\ /\
a j # #
/\ /\
## i#
/\
##
What will we print?
From k, we go left to
h, where we go left to
a, where we go left to
null, where we do nothing and so
we return to a, where we print 'a' and then go right to
null, where we do nothing and so
we return to a and are done, so
we return to h, where we print 'h' and then go right to
j, where we go left to
i, where we go left to
null, where we do nothing and so
we return to i, where we print 'i' and then go right to
null, where we do nothing and so
we return to i and are done, so
we return to j, where we print 'j' and then go right to
null, where we do nothing and so
we return to j and are done, so
we return to h and are done, so
we return to k, where we print 'k' and then go right to
n where we go left to
null, where we do nothing and so
we return to n, where we print 'n' and then go right to
null, where we do nothing and so
we return to n and are done, so
we return to k and are done, so we return to the caller
So if we just look at the lines were we printed:
we return to a, where we print 'a' and then go right to
we return to h, where we print 'h' and then go right to
we return to i, where we print 'i' and then go right to
we return to j, where we print 'j' and then go right to
we return to k, where we print 'k' and then go right to
we return to n, where we print 'n' and then go right to
We see we printed "ahijkn", which is indeed in alphabetical order.
We manage to print an entire tree, in alphabetical order, just by knowing how to print a single node in alphabetical order. Which was just (because our tree had the special property of ordering values to the left of alphabetically later values) knowing to print the left child before printing the node's value, and to print the right child after printing the node's value.
And that's the power of recursion: being able to do whole things by knowing only how to do a part of the whole (and knowing when to stop recursing).
Recalling that in most languages, operator || ("or") short-circuits when its first operand is true, the general recursive function is:
void recurse() { doWeStop() || recurse(); }
Luc M comments:
SO should create a badge for this kind of answer. Congratulations!
Thanks, Luc! But, actually, because I edited this answer more than four times (to add the last example, but mostly to correct typos and polish it -- typing on a tiny netbook keyboard is hard), I can't get any more points for it. Which somewhat discourages me from putting as much effort into future answers.
See my comment here on that: https://stackoverflow.com/questions/128434/what-are-community-wiki-posts-in-stackoverflow/718699#718699
Your brain blew up because it got into an infinite recursion. That's a common beginner mistake.
Believe it or not, you already understand recursion, you're just being dragged down by a common, but faulty metaphor for a function: a small box with stuff that comes in and out.
Think instead of a task or procedure, such as "find out more about recursion on the net". That's recursive and you have no problem with it. To complete this task you might:
a) Read a Google's result page for "recursion"
b) Once you've read it, follow the first link on it and...
a.1)Read that new page about recursion
b.1)Once you've read it, follow the first link on it and...
a.2)Read that new page about recursion
b.2)Once you've read it, follow the first link on it and...
As you can see, you've been doing recursive stuff for a long time without any problems.
For how long would you keep doing that task? Forever until your brain blows up? Of course not, you will stop at a given point, whenever you believe you have completed the task.
There's no need to specify this when asking you to "find out more about recursion on the net", because you are a human and you can infer that by yourself.
Computers can't infer jack, so you must include an explicit ending: "find out more about recursion on the net, UNTIL you understand it or you have read a maximum of 10 pages".
You also inferred that you should start at Google's result page for "recursion", and again that's something a computer can't do. The complete description of our recursive task must also include an explicit starting point:
"find out more about recursion on the net, UNTIL you understand it or you have read a maximum of 10 pages and starting at www.google.com/search?q=recursion"
To grok the whole thing, I suggest you try any of these books:
Common Lisp: A Gentle Introduction to Symbolic Computation. This is the cutest non-mathematical explanation of recursion.
The little schemer.
To understand recursion, all you have to do is look on the label of your shampoo bottle:
function repeat()
{
rinse();
lather();
repeat();
}
The problem with this is that there is no termination condition, and the recursion will repeat indefinitely, or until you run out of shampoo or hot water (external termination conditions, similar to blowing your stack).
If you want a book that does a good job of explaining recursion in simple terms, take a look at Gödel, Escher, Bach: An Eternal Golden Braid by Douglas Hofstadter, specifically Chapter 5. In addition to recursion it does a nice job of explaining a number of complex concepts in computer science and math in an understandable way, with one explanation building on another. If you haven't had much exposure to these sorts of concepts before, it can be a pretty mindblowing book.
This is more of a complaint than a question. Do you have a more specific question on recursion? Like multiplication, it's not a thing people write a lot about.
Speaking of multiplication, think of this.
Question:
What's a*b?
Answer:
If b is 1, it's a.
Otherwise, it's a+a*(b-1).
What's a*(b-1)? See the above question for a way to work it out.
Actually you use recursion to reduce the complexity of your problem at hand. You apply recursion until you reach a simple base case that can be solved easily. With this you can solve the last recursive step. And with this all other recursive steps up to your original problem.
I think this very simple method should help you understand recursion. The method will call itself until a certain condition is true and then return:
function writeNumbers( aNumber ){
write(aNumber);
if( aNumber > 0 ){
writeNumbers( aNumber - 1 );
}
else{
return;
}
}
This function will print out all numbers from the first number you'll feed it till 0. Thus:
writeNumbers( 10 );
//This wil write: 10 9 8 7 6 5 4 3 2 1 0
//and then stop because aNumber is no longer larger then 0
What bassicly happens is that writeNumbers(10) will write 10 and then call writeNumbers(9) which will write 9 and then call writeNumber(8) etc. Until writeNumbers(1) writes 1 and then calls writeNumbers(0) which will write 0 butt will not call writeNumbers(-1);
This code is essentially the same as:
for(i=10; i>0; i--){
write(i);
}
Then why use recursion you might ask, if a for-loop does essentially the same. Well you mostly use recursion when you would have to nest for loops but won't know how deep they are nested. For example when printing out items from nested arrays:
var nestedArray = Array('Im a string',
Array('Im a string nested in an array', 'me too!'),
'Im a string again',
Array('More nesting!',
Array('nested even more!')
),
'Im the last string');
function printArrayItems( stringOrArray ){
if(typeof stringOrArray === 'Array'){
for(i=0; i<stringOrArray.length; i++){
printArrayItems( stringOrArray[i] );
}
}
else{
write( stringOrArray );
}
}
printArrayItems( stringOrArray );
//this will write:
//'Im a string' 'Im a string nested in an array' 'me too' 'Im a string again'
//'More nesting' 'Nested even more' 'Im the last string'
This function could take an array which could be nested into a 100 levels, while you writing a for loop would then require you to nest it 100 times:
for(i=0; i<nestedArray.length; i++){
if(typeof nestedArray[i] == 'Array'){
for(a=0; i<nestedArray[i].length; a++){
if(typeof nestedArray[i][a] == 'Array'){
for(b=0; b<nestedArray[i][a].length; b++){
//This would be enough for the nestedAaray we have now, but you would have
//to nest the for loops even more if you would nest the array another level
write( nestedArray[i][a][b] );
}//end for b
}//endif typeod nestedArray[i][a] == 'Array'
else{ write( nestedArray[i][a] ); }
}//end for a
}//endif typeod nestedArray[i] == 'Array'
else{ write( nestedArray[i] ); }
}//end for i
As you can see the recursive method is a lot better.
I'll try to explain it with an example.
You know what n! means? If not: http://en.wikipedia.org/wiki/Factorial
3! = 1 * 2 * 3 = 6
here goes some pseudocode
function factorial(n) {
if (n==0) return 1
else return (n * factorial(n-1))
}
So let's try it:
factorial(3)
is n 0?
no!
so we dig deeper with our recursion:
3 * factorial(3-1)
3-1 = 2
is 2 == 0?
no!
so we go deeper!
3 * 2 * factorial(2-1)
2-1 = 1
is 1 == 0?
no!
so we go deeper!
3 * 2 * 1 * factorial(1-1)
1-1 = 0
is 0 == 0?
yes!
we have a trivial case
so we have
3 * 2 * 1 * 1 = 6
i hope the helped you
Recursion
Method A, calls Method A calls Method A. Eventually one of these method A's won't call and exit, but it's recursion because something calls itself.
Example of recursion where I want to print out every folder name on the hard drive: (in c#)
public void PrintFolderNames(DirectoryInfo directory)
{
Console.WriteLine(directory.Name);
DirectoryInfo[] children = directory.GetDirectories();
foreach(var child in children)
{
PrintFolderNames(child); // See we call ourself here...
}
}
A recursive function is simply a function that calls itself as many times as it needs to do so. It's useful if you need to process something multiple times, but you're unsure how many times will actually be required. In a way, you could think of a recursive function as a type of loop. Like a loop, however, you'll need to specify conditions for the process to be broken otherwise it'll become infinite.
Which book are you using?
The standard textbook on algorithms that is actually good is Cormen & Rivest. My experience is that it teaches recursion quite well.
Recursion is one of the harder parts of programming to grasp, and while it does require instinct, it can be learned. But it does need a good description, good examples, and good illustrations.
Also, 30 pages in general is a lot, 30 pages in a single programming language is confusing. Don't try to learn recursion in C or Java, before you understand recursion in general from a general book.
http://javabat.com is a fun and exciting place to practice recursion. Their examples start fairly light and work through extensive (if you want to take it that far). Note: Their approach is learn by practicing. Here is a recursive function that I wrote to simply replace a for loop.
The for loop:
public printBar(length)
{
String holder = "";
for (int index = 0; i < length; i++)
{
holder += "*"
}
return holder;
}
Here is the recursion to do the same thing. (notice we overload the first method to make sure it is used just like above). We also have another method to maintain our index (similar to the way the for statement does it for you above). The recursive function must maintain their own index.
public String printBar(int Length) // Method, to call the recursive function
{
printBar(length, 0);
}
public String printBar(int length, int index) //Overloaded recursive method
{
// To get a better idea of how this works without a for loop
// you can also replace this if/else with the for loop and
// operationally, it should do the same thing.
if (index >= length)
return "";
else
return "*" + printBar(length, index + 1); // Make recursive call
}
To make a long story short, recursion is a good way to write less code. In the latter printBar notice that we have an if statement. IF our condition has been reached, we will exit the recursion and return to the previous method, which returns to the previous method, etc. If I sent in a printBar(8), I get ********. I am hoping that with an example of a simple function that does the same thing as a for loop that maybe this will help. You can practice this more at Java Bat though.
The truly mathematical way to look at building a recursive function would be as follows:
1: Imagine you have a function that is correct for f(n-1), build f such that f(n) is correct.
2: Build f, such that f(1) is correct.
This is how you can prove that the function is correct, mathematically, and it's called Induction. It is equivalent to have different base cases, or more complicated functions on multiple variables). It is also equivalent to imagine that f(x) is correct for all x
Now for a "simple" example. Build a function that can determine if it is possible to have a coin combination of 5 cents and 7 cents to make x cents. For example, it's possible to have 17 cents by 2x5 + 1x7, but impossible to have 16 cents.
Now imagine you have a function that tells you if it's possible to create x cents, as long as x < n. Call this function can_create_coins_small. It should be fairly simple to imagine how to make the function for n. Now build your function:
bool can_create_coins(int n)
{
if (n >= 7 && can_create_coins_small(n-7))
return true;
else if (n >= 5 && can_create_coins_small(n-5))
return true;
else
return false;
}
The trick here is to realize that the fact that can_create_coins works for n, means that you can substitute can_create_coins for can_create_coins_small, giving:
bool can_create_coins(int n)
{
if (n >= 7 && can_create_coins(n-7))
return true;
else if (n >= 5 && can_create_coins(n-5))
return true;
else
return false;
}
One last thing to do is to have a base case to stop infinite recursion. Note that if you are trying to create 0 cents, then that is possible by having no coins. Adding this condition gives:
bool can_create_coins(int n)
{
if (n == 0)
return true;
else if (n >= 7 && can_create_coins(n-7))
return true;
else if (n >= 5 && can_create_coins(n-5))
return true;
else
return false;
}
It can be proven that this function will always return, using a method called infinite descent, but that isn't necessary here. You can imagine that f(n) only calls lower values of n, and will always eventually reach 0.
To use this information to solve your Tower of Hanoi problem, I think the trick is to assume you have a function to move n-1 tablets from a to b (for any a/b), trying to move n tables from a to b.
Simple recursive example in Common Lisp:
MYMAP applies a function to each element in a list.
1) an empty list has no element, so we return the empty list - () and NIL both are the empty list.
2) apply the function to the first list, call MYMAP for the rest of the list (the recursive call) and combine both results into a new list.
(DEFUN MYMAP (FUNCTION LIST)
(IF (NULL LIST)
()
(CONS (FUNCALL FUNCTION (FIRST LIST))
(MYMAP FUNCTION (REST LIST)))))
Let's watch the traced execution. On ENTERing a function, the arguments are printed. On EXITing a function, the result is printed. For each recursive call, the output will be indented on level.
This example calls the SIN function on each number in a list (1 2 3 4).
Command: (mymap 'sin '(1 2 3 4))
1 Enter MYMAP SIN (1 2 3 4)
| 2 Enter MYMAP SIN (2 3 4)
| 3 Enter MYMAP SIN (3 4)
| | 4 Enter MYMAP SIN (4)
| | 5 Enter MYMAP SIN NIL
| | 5 Exit MYMAP NIL
| | 4 Exit MYMAP (-0.75680256)
| 3 Exit MYMAP (0.14112002 -0.75680256)
| 2 Exit MYMAP (0.9092975 0.14112002 -0.75680256)
1 Exit MYMAP (0.841471 0.9092975 0.14112002 -0.75680256)
This is our result:
(0.841471 0.9092975 0.14112002 -0.75680256)
To explain recursion to a six-year-old, first explain it to a five-year-old, and then wait a year.
Actually, this is a useful counter-example, because your recursive call should be simpler, not harder. It would be even harder to explain recursion to a five-year old, and though you could stop the recursion at 0, you have no simple solution for explaining recursion to a zero-year-old.
To solve a problem using recursion, first sub-divide it into one or more simpler problems that you can solve in the same way, and then when the problem is simple enough to solve without further recursion, you can return back up to higher levels.
In fact, that was a recursive definition of how to solve a problem with recursion.
Children implicitly use recursion, for instance:
Road trip to Disney World
Are we there yet?(no)
Are we there yet?(Soon)
Are we there yet?(Almost...)
Are we there yet?(SHHHH)
Are we there yet?(!!!!!)
At which point the child falls asleep...
This countdown function is a simple example:
function countdown()
{
return (arguments[0] > 0 ?
(
console.log(arguments[0]),countdown(arguments[0] - 1)) :
"done"
);
}
countdown(10);
Hofstadter's Law applied to software projects is also relevant.
The essence of human language is, according to Chomsky, the ability of finite brains to produce what he considers to be infinite grammars. By this he means not only that there is no upper limit on what we can say, but that there is no upper limit on the number of sentences our language has, there's no upper limit on the size of any particular sentence. Chomsky has claimed that the fundamental tool that underlies all of this creativity of human language is recursion: the ability for one phrase to reoccur inside another phrase of the same type. If I say "John's brother's house", I have a noun, "house", which occurs in a noun phrase, "brother's house", and that noun phrase occurs in another noun phrase, "John's brother's house". This makes a lot of sense, and it's an interesting property of human language.
References
Recursion and Human Thought
When working with recursive solutions, I always try to:
Establish the base case first i.e.
when n = 1 in a solution to factorial
Try to come up with a general rule
for every other case
Also there are different types of recursive solutions, there's the divide and conquer approach which is useful for fractals and many others.
It would also help if you could work on simpler problems first just to get the hang of it. Some examples are solving for the factorial and generating the nth fibonacci number.
For references, I highly recommend Algorithms by Robert Sedgewick.
Hope that helps. Good luck.
A recursive function is like a spring you compress a bit on each call. On each step, you put a bit of information (current context) on a stack. When the final step is reached, the spring is released, collecting all values (contexts) at once!
Not sure this metaphor is effective... :-)
Anyway, beyond the classical examples (factorial which is the worst example since it is inefficient and easily flattened, Fibonacci, Hanoi...) which are a bit artificial (I rarely, if ever, use them in real programming cases), it is interesting to see where it is really used.
A very common case is to walk a tree (or a graph, but trees are more common, in general).
For example, a folder hierarchy: to list the files, you iterate on them. If you find a sub-directory, the function listing the files call itself with the new folder as argument. When coming back from listing this new folder (and its sub-folders!), it resumes its context, to the next file (or folder).
Another concrete case is when drawing a hierarchy of GUI components: it is common to have containers, like panes, to hold components which can be panes too, or compound components, etc. The painting routine calls recursively the paint function of each component, which calls the paint function of all the components it holds, etc.
Not sure if I am very clear, but I like to show real world use of teaching material, as it was something I was stumbling upon in the past.
Ouch. I tried to figure out the Towers of Hanoi last year. The tricky thing about TOH is it's not a simple example of recursion - you have nested recursions which also change the roles of towers on each call. The only way I could get it to make sense was to literally visualize the movement of the rings in my mind's eye, and verbalize what the recursive call would be. I would start with a single ring, then two, then three. I actually ordered the game on the internet. It took me maybe two or three days of cracking my brains to get it.
Think a worker bee. It tries to make honey. It does its job and expects other worker bees to make rest of the honey. And when the honeycomb is full, it stops.
Think it as magic. You have a function that has the same name with the one you are trying to implement and when you give it the subproblem, it solves it for you and the only thing you need to do is to integrate the solution of your part with the solution it gave you.
For example, we want to calculate the length of a list. Lets call our function magical_length and our magical helper with magical_length
We know that if we give the sublist which does not have the first element, it will give us the length of the sublist by magic. Then only thing we need to think is how to integrate this information with our job. The length of the first element is 1 and magic_counter gives us the length of sublist n-1, therefore total length is (n-1) + 1 -> n
int magical_length( list )
sublist = rest_of_the_list( list )
sublist_length = magical_length( sublist ) // you can think this function as magical and given to you
return 1 + sublist_length
However this answer is incomplete because we didn't consider what happens if we give an empty list. We thought that the list we have always has at least one element. Therefore we need to think about what should be the answer if we are given an empty list and answer is obviously 0. So add this information to our function and this is called base/edge condition.
int magical_length( list )
if ( list is empty) then
return 0
else
sublist_length = magical_length( sublist ) // you can think this function as magical and given to you
return 1 + sublist_length