Pascal's Formula Recusrion (not triangle) Announcing the arrival of Valued Associate #679:...
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Pascal's Formula Recusrion (not triangle)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Redesign a module to be genericCritique of Pascal's Triangle ClassCondensing code that generates Pascal's trianglePerceptron algorithmBlocking reads when writes are happening on two flowsMultithreaded Pascal's Triangle CalculatorCalculate Kth Row of Pascal's TriangleCompare 2 unordered, rooted trees for shape-isomorphismFollow-up 1: Compare 2 unordered, rooted trees for shape-isomorphismPascal's Triangle - Java Recursion
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So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect
static void printSubsets(int []B, int k, int i){
//int n = 0;
if(k==0) {
System.out.println(setToString(B));
}
if(k > B.length - i) {
return ;
}
else if (B[i]!=){
printSubsets(B, k, i+1);
}
else if (B[i]==1){
printSubsets(B, k-1, i+1);;
}
}
java
asked 6 mins ago
EternityEternity
1
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect
static void printSubsets(int []B, int k, int i){
//int n = 0;
if(k==0) {
System.out.println(setToString(B));
}
if(k > B.length - i) {
return ;
}
else if (B[i]!=){
printSubsets(B, k, i+1);
}
else if (B[i]==1){
printSubsets(B, k-1, i+1);;
}
}
java
asked 6 mins ago
EternityEternity
1
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect
static void printSubsets(int []B, int k, int i){
//int n = 0;
if(k==0) {
System.out.println(setToString(B));
}
if(k > B.length - i) {
return ;
}
else if (B[i]!=){
printSubsets(B, k, i+1);
}
else if (B[i]==1){
printSubsets(B, k-1, i+1);;
}
}
java
asked 6 mins ago
EternityEternity
1
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect
static void printSubsets(int []B, int k, int i){
//int n = 0;
if(k==0) {
System.out.println(setToString(B));
}
if(k > B.length - i) {
return ;
}
else if (B[i]!=){
printSubsets(B, k, i+1);
}
else if (B[i]==1){
printSubsets(B, k-1, i+1);;
}
}
java
java
asked 6 mins ago
EternityEternity
1
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
EternityEternity
1
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
EternityEternity
1
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
EternityEternity
1
asked 6 mins ago
EternityEternity
1
1
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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