Originally published at: http://boingboing.net/2017/08/11/martin-gardner-puzzle-red-wh.html

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The balls are R1, R2, W1, W2, B1, B2.

put R1 and W2 on the left side of the scale, W1 and B2 on the right side.

If it goes left, R1 and W2 are the heavy ones, compare one of these to the blue and you’ve got them all.

If it goes right, W1 and B2 are the heavy ones, compare one of these to the blue and you’ve got them all.

If it balances, You already know that W1 and W2 are opposite, so R1 is opposite of B2 the other way, and R2 must be opposite B1, so weigh any 1 of these pairs against each other, to see which is heavier and you can show them all:

W1 vs. W2:

If W1 is heavier:

W2 must be lighter

B2 must be lighter

B1 must be heavier

R1 must be heavier

R2 must be lighter

If W2 is heavier, opposite as above.

Left Side:1 each of red, white, and blue balls

Right Side: the other red, white, and blue balls

Discard the three heavier balls.

Weigh two of the remaining three balls. If they weigh the same, the unweighed ball is lighter. Otherwise, the lighter ball will be obvious on the scale.

Edit: I didn’t read the information carefully and I fucked up. My way won’t work.

It may be out of **print**, but it is still available, along with, i believe, all of his books that were from his *Scientific American* columns in digital format:

https://www.maa.org/press/books/martin-gardner-s-mathematical-games-the-entire-collection-of-his-scientific-american-columns-on-one

Worked it out in Excel: there are 8 weight options for red, blue and white (0 = light, 1 = heavy). Test results are indicated by color (left = green, right = red, balanced = blue).

Test 1: R1 W2 vs. B1 W1

Test 2: R1 B1 vs. B2 W1

Row that matches the test results for T1 & T2 gives the weights.

I think you are making a logical leap here. In your first weighing, if R1 and B2 are both the lighter ones or the heavier ones, the scale will just go to whichever side the heavy W is from. If we then compare the R1 to the B2 they will be equal and that doesn’t tell us anything we don’t already know.

My go at it:

[spoiler]One of each color on each side. Take the heavier side, and weigh two of those. Whichever is heavier is the heavier of its pair, and the lighter is the lighter, and the third from that side is also heavier.

If they are equal you are screwed, but the odds of that are only 1/3 so it’s a safe enough risk.

edit: actual 50% chance of failure. This is hard.[/spoiler]

Place both blue balls on one side (one heavy and one light per the description). Place one red and one white on the other side. There are three possibilities:

- Red/white side is heavier. Both must be heavy balls, therefore other red and white balls are light. Use the second weighing one blue ball against the other, and all weights are determined.
- Red/white side is lighter, meaning both balls are the light ones and the remaining are heavy. Weigh blue balls as in option one for the second cycle and all weights are determined.
- Weights are the same. Because you know that the blue balls are one heavy and one light, the same is true of the red and white balls, but you do not know which is which. Use your second weighing by switching the red ball with one blue ball. Two possibilities exist:

1a- the red and blue balls are heavier. Therefore, the white ball on the scale is light, the other blue ball is light, and the unweighed red ball is light. The unweighed white ball is heavy.

2a- the blue and white balls are heavier. That means the other blue ball is light, the red weighed red ball is light and unweighed is heavy, and the remaining unweighed white ball is light.

Think I got it!

There’s another option, 3. 3a - the blue and white ball are the same weight as the red and blue ball. Either the red ball is heavy and the white ball light, or the other way around, you cannot tell.

I went down this path too but I think there should be a 3a - weights are the same. When you switched the red ball with a blue ball, the blue ball selected may weigh the same as the replaced red ball and we’re still stuck I think.

One other possibility in the second weighing you switch one heavy for the other heavy one light for the other light so the weights are the same in the second weighing, so you can’t deconvolute which is heavy and which is light. Thought I had it! So here is the solution I think works.

On first weigh, weigh one blue and one white on one side, one red and one blue on the other. If one side is heavier, the blue ball must be the heavier one on that side. the red and white balls can either be the same weight (both light or both heavy) or the red and white could be different (heavy on the heavy blue side and light on the light blue side). On the second weigh (let’s say the red ball was on the heavy side but it is equally valid if it is white one), switch the red ball with the unweighed one, and weigh against the white ball from the first round. Three possibilities occur:

- The red weighs less than the white. This means you switched from the heavy red ball from the first round, which is consistent with the first round being heavier and the original red and white balls weighing the same (heavy).
- The red weighs more than the white. This means you switched from the light red ball from the first round, which is consistent with the first round being heavier and the original red and white balls weighing the same (light).
- The red weighs the same as the white. This means you switched a heavy red ball for a light one, and this is consistent with case 3 in the first weigh where the heavy red ball is on the same side as a heavy blue ball, and the opposite side had a light white ball and light blue ball.

Obviously I am saying that the red ball was on the heavy side in this case, but the logic holds if it was the white ball instead.

If the weights on the first weigh are the same, you know that one of the red/ white balls are heavy and the other is light. Switch the blue balls as you now have to have two heavy balls on one side and two light on the other, and you know all the weights! Pretty sure I got it for real this time!

I thought I solved this twice and figured out while I was writing the comments that I was wrong. *thinks some more*

This is the answer. You can prove with a flow chart.

Weighing One:

On the left side place one red ball, one white ball. On the right side, place one red ball and one blue ball.

Case1 (the easy case): the scale is not balanced (assume the right side is heavier). This means the right side has two heavy balls. You now know the right side has the Heavy Red Ball and the Heavy Blue Ball, and the left side is Light Red Ball and Light White Ball. The unused balls are Light Blue Ball and Heavy White Ball. DONE.

Case 2: the scale is balanced. This means each side has one heavy and one light ball. Since both red balls are on the scale and are opposite weights, you know the white and blue balls that are on the scale are also opposite weights.

Remove the balls, but remember which blue and white balls were weighed, and the position of the red balls in Weighing One.

Weighing Two:

Left Side place one blue ball; Right Side, place the other blue ball. You know which blue ball was weighed in Weighing One.

The scale will be unbalanced and you will immediately know which blue ball is heavy and which is light. Assume the heavy side is the Right Side (Heavy Blue Ball on the Right).

If the Right Side is the ball that was weighed in Weighing One, then you know that the Right Side in Weighing One contained a Light Red Ball and the Left Side in Weighing One contained a Light White Ball and a Heavy Red Ball. The unused White Ball is Heavy. You’re Done.

If the Right Side Ball is NOT the ball that was weighed in Weighing One, then the Left Side Light Blue Ball was used in Weighing One. This means, in Weighing One, the Right Side had a Light Blue Ball and a Heavy Red Ball, and the Weighing One Left Side was a Light Red Ball and a Heavy White Ball. The unused white ball is Light. You’re Done.

This works! f the weights on the first weigh are the same, you could also weight only the red & white from the first weighing against each other (without the blues).

The scale is not balanced with white & red heavy and white & red light as well, so after this you’ll only know the weight of the blue balls - you’ll need a second test.

I didn’t realise the second test (in the case of a balanced first result) could be done with the blue balls as well, but of course you’re right on this.

Also, I’m saying blue balls a lot.

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