Brain Teaser: Solve Chocolate Bar Puzzle in Min Breaks

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SUMMARY

The Chocolate Bar Puzzle involves determining the minimum number of breaks required to split a chocolate bar measuring 6 blocks wide by 9 blocks long into individual blocks. The established formula for the minimum number of breaks is w + m - 2, where w represents the width and m represents the length of the bar. For the given dimensions, this results in 13 breaks, which is the upper limit. The discussion also clarifies that to achieve this, one must make (w-1) + (m-1) breaks, totaling 13 for the specified dimensions.

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fernanroy
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A friend of mine is really pulling my chain. He said the answer to this was right in front of me. I doubt that! Does anyone have an idea on this?


"I have a bar of chocolate 6 blocks wide by 9 blocks long and I want to split it into its individual blocks making the smallest number of breaks that I can. The breaks that I am allowed to make start on one side and finish on a different side without visiting a third side.


How many breaks must I make?

Can you generalize your answer bard that are to w blocks wide and m blocks long?"
 
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Let's see. The obvious answer is 13, or w+m-2. I'm sure there's a shorter answer. That's the upper limit.
 
Thank You!

DaveC426913 said:
Let's see. The obvious answer is 13, or w+m-2. I'm sure there's a shorter answer. That's the upper limit.

Makes sense now that I look at it! Would you elaborate w+m-2?
 
fernanroy said:
Would you elaborate w+m-2?
To break something into 6 pieces, you to need to make 5 breaks.

So, specifically: 9-1 + 6-1, or generally: w-1 + m-1.
 
DaveC426913 said:
To break something into 6 pieces, you to need to make 5 breaks.

So, specifically: 9-1 + 6-1, or generally: w-1 + m-1.

YES! Thanks so much! I appreciate it!
 
Hmm, what about breaking it then stacking it and breaking it again? How thick is this bar?
 

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