checkpoint1 #8
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reneemeyer
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Nov 6, 2018
reneemeyer
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reneemeyer
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Good job, just some function organization changes. Overall though, great methods
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| function movePiece(startStack, endStack) { | ||
| if (startStack === 'a' && endStack === 'b') { | ||
| let mover = stacks.a.pop(); |
| // Your code here | ||
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| function movePiece(startStack, endStack) { | ||
| if (startStack === 'a' && endStack === 'b') { |
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for every stack and letter, you're checking if the stack is a,b, or c, twice, that's a lot of repetition that you can condense.
| let mover = stacks.a.pop(); | ||
| stacks.b.push(mover); | ||
| } else if (startStack === 'a' && endStack === 'c') { | ||
| let mover = stacks.a.pop(); |
| let mover = stacks.a.pop(); | ||
| stacks.c.push(mover); | ||
| } else if (startStack === 'b' && endStack === 'a') { | ||
| let mover = stacks.b.pop(); |
| let mover = stacks.b.pop(); | ||
| stacks.a.push(mover); | ||
| } else if (startStack === 'b' && endStack === 'c') { | ||
| let mover = stacks.b.pop(); |
| let mover = stacks.b.pop(); | ||
| stacks.c.push(mover); | ||
| } else if (startStack === 'c' && endStack === 'a') { | ||
| let mover = stacks.c.pop(); |
| let mover = stacks.c.pop(); | ||
| stacks.a.push(mover); | ||
| } else { | ||
| let mover = stacks.c.pop(); |
| function isLegal(start, end) { | ||
| const startStack = stacks[start]; | ||
| const endStack = stacks[end]; | ||
| if (startStack.length === 0) { |
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anytime that you see that you're returning true or false, return the evaluation
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| function checkForWin() { | ||
| // Your code here | ||
| if (stacks.b.length === 4 || stacks.c.length === 4) { |
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anytime that you see that you're returning true or false, return the evaluation
| function towersOfHanoi(startStack, endStack) { | ||
| // Your code here | ||
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| movePiece(startStack, endStack); |
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you should only move the piece if your function is legal, and you should only check for win if your function is legal
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