This was a simple puzzle and many of you sent me correct answers. Here are the first 5 to respond, each of whom have been sent an Amazon gift coupon.

Gourav Sharma – 6.45 pm Oct 29

Suman Saraf – 6.49 pm Oct 29

Deepak Kumar Singh – 6.56 pm Oct 29

Kartik Aggarwal – 11.38 pm Oct 29

Abdul Hakeem – 6.09 am Oct 30

I am copying the answer from one of the respondents. Thanks again to my daughter, Arushi Sharma Goyal, for contributing this puzzle.

Hope you all enjoyed the puzzle!

]]>Many people sent correct answers – this includes Suman Saraf, Kumz91 (sorry do not know the full name), Naresh Kumra, Mohit Rana, Delhi Scrabble (again, do not know the real name), Gaurav Sharma, Viv423 (do not know the name again), Rishikant, G, Rowan an Pratik Poddar. And sorry, if I have missed out anyone. Well done all.

Please note that each cube must have 0, 1 and 2. This leaves only six faces for the 7 remaining digits. Trick is to realize that 6 and 9 can be on the same face and the cube can be turned around. On the right one, therefore, the three hidden faces will be 0, 1 and 2. On the left cube, one can see 1 and 2. The others will be 0, 6/9, 7 and 8.

Hope you all enjoyed the puzzle!

]]>This puzzle was given to be last week by my younger daughter, who got this in her class (grade V) in a math quiz. Its a simple puzzle, where you need to find your way from the entry to the GOAL without taking any left turn.

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy solving!

]]>The site from where I took this puzzle had many correct answers, most of these were largely a bit of trial and error. One solution, which I modified a little, is as follows:

For each of the output node, if the desired input for firing is X, Y, Z (each of these being either a 0 or a 1), design the weights as:

- -1 if the input should be 0
- 1 if the input should have been 1

When you do this, you get an input output matrix as follows:

If we keep the threshold to be just below the thresholds mentioned in the table, you will get the desires output.

Hope you all enjoyed the solution!

]]>Please look at the figure above. This is a desk calendar made out of two cubes that Martin Gardner saw at the Grand Central station in New York in a shop. Each day from 01-31 can be indicated by the two cubes so that their front faces give the required date. Each face can have a unique digit from 0 through 9. What are the four digits that cannot be seen on the left cube and the three that cannot be seen on the right cube?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy dating!

]]>The answer is 1024.

An easy way to think about this is to create a 10 digit binary number, where each digit represents one of the friends each. 0 means the friend is not coming, and 1 means that the friend is coming. Note that there can be 2^10 combinations here, and each of them represents a distinct combination of friends coming.

Hope you all enjoyed the puzzle!

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Neural networks and deep learning are the buzzwords of the day. Here is a puzzle that shows the basic mechanics of a neural network.

We’re going to create a simple network that converts binary numbers to decimal numbers. Imagine a network with just two layers — an input layer consisting of three units and an output layer with seven units. Each unit in the first layer connects to each unit in the second, as shown in the figure below.

As you can see, there are 21 connections. Every unit in the input layer, at a given point in time, is either off or on, having a value of 0 or 1. Every connection from the input layer to the output layer has an associated weight that, in artificial neural networks, is a real number between 0 and 1. Just to be contrary, and to make the network a touch more similar to an actual neural network, let us allow the weight to be a real number between –1 and 1 (the negative sign signifying an inhibitory neuron). The product of the input’s value and the connection’s weight is conveyed to an output unit as shown below. The output unit adds up all the numbers it gets from its connections to obtain a single number, as shown in the figure below using arbitrary input values and connection weights for a single output unit. Based on this number, the output number decides what its state is going to be. If it is more than a certain threshold, the unit’s value becomes 1, and if not, then its value becomes 0. We can call a unit that has a value of 1 a “firing” or “activated” unit and a unit with a value of 0 a “quiescent” unit.

Our three input units from top to bottom can have the values 001, 010, 011, 100, 101, 110 or 111, which readers may recognize as the binary numbers from 1 through 7.

Now the question: Is it possible to adjust the connection weights and the thresholds of the seven output units such that every binary number input results in the firing of only one appropriate output unit, with all the others being quiescent? The position of the activated output unit should reflect the binary input value. Thus, if the original input is 001, the leftmost output unit (or bottommost when viewed on a phone, as shown in the top figure) alone should fire, whereas if the original input is 101, the output unit that is fifth from the left (or from the bottom on a phone) alone should fire, and so on.

If you think the above result is not possible, try to adjust the weights to get as close as you can. Can you think of a simple adjustment, using additional connections but not additional units, that can improve your answer?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy neuron-firing over the weekend!

]]>This was not a very difficult puzzle, but I got correct answers only from the two people who solve every puzzle – Pratik Poddar and Suman Saraf – thank you and well done!

(a) Color the board like a chessboard, i.e. alternate black and whites. Notice that a bug on a white square can only travel to a black and vice versa. Since there are a total of 81 squares, there will be one more white square (or one more black square). Hence when the bugs move, there will correspondingly be at least one empty white square (or black square).

(b) Bugs can change places pairwise. One can leave only the bug in the centre square or the corner, which will be the only empty square.

Hope you all enjoyed the puzzle!

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A person has 10 friends. Over several days he invites some of them to a dinner party so that the company never repeats itself i.e. the exactly the same set of people cannot repeat itself on any of the days. He may, however, not invite anyone on one of the days. For how many days can he continue to invite people without repetition?

Happy partying…and there will be a lot of it!

]]>This puzzle can be solved using the concept of “invariants” and parity.

Note that for any numbers a and b (a > b), numbers a and b get replaced by c. If a and b are both odd or both even, then c will be even. If one of them is odd and the other is even, then c will be odd. Essentially, in all the cases, if (a+b) is odd, c will be odd; If (a+b) is even, then c will be even. Therefore, parity of a+b is an invariant.

For the final set of numbers to be zero (which is even parity), the sum of all the numbers must be even, i.e. 1, 2, 3, …., 1989 must add up to an even number, which is not the case. Hence we can never achieve zeros as the final state.

Hope you all enjoyed the puzzle!

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