Reading from a file in tcl

It is very common to read from and/or write to a file in any programming language. In tcl also, one frequently uses file operations. read command in tcl reads the entire file and stores it into a variable. One can, then, perform the desired operations onto the read data. The normal command sequence to read a file in tcl language is as below:

// script to read and display contents

set infile [open input_file.rpt r] // Create a file pointer and point it to the file to be read

set file_data [read $infile]            // Assign file_data with contents of infile

close $infile                                        // Detach the file pointer from file to be read

set lines [split $file_data “\n”]   

// Split the file contents by lines and assign each line to an element of list

foreach element $lines {

                puts $element                  // Display each element of $lines onto screen.

}

2-input AND gate using 2:1 mux

2-input AND gate implementation using 2:1 mux: Figure 1 below shows the truth table of a 2-input AND gate. If we observe carefully, OUT equals '0' when A is '0'. And OUT follows B when A is '1'. So, if we connect A to the select pin of a 2:1 mux, AND gate will be implemented if we connect D0 to '0' and D1 to 'B'.

Figure 1: Truth table of AND gate

Figure 2 below shows the implementation of 2-input AND gate using a 2:1 multiplexer.

Figure 2: Implementation of AND gate using a 2:1 mux

Also read:

• NOT gate using 2:1 mux
• 2-input XOR gate using 2:1 mux
• 2-input NOR gate using 2:1 mux
• 2-input XNOR gate using 2:1 mux
• 2-input OR gate using 2:1 mux
• 2-input NAND gate using 2:1 mux

2-input NAND gate using 2:1 mux

2-input NAND gate using 2:1 mux: Figure 3 below shows the truth table of a 2-input NAND gate. If we observe carefully, OUT equals '1' when A is '0'. Similarly, when A is '1', OUT is B'. So, if we connect SEL pin of mux to A, D0 pin of mux to '1' and D1 to B', then it will act as a NAND gate.

Figure 3: Truth table of 2-input NAND gate

Figure 4 below shows the implementation of a 2-input NAND gate using 2:1 mux.

Figure 4: Implementation of 2-input NAND gate using 2:1 mux

Also read:

• NOT gate using 2:1 mux
• 2-input XOR gate using 2:1 mux
• 2-input NOR gate using 2:1 mux
• 2-input XNOR gate using 2:1 mux
• 2-input OR gate using 2:1 mux
• 2-input AND gate using 2:1 mux

2-input OR gate using 2:1 mux

2-input OR gate using 2x1 mux: Figure 5 below shows the truth table for a 2-input OR gate. If we observe carefully, OUT equals B when A is '0'. Similarly, OUT is '1' (or A), when A is '1'. So, we can make a 2:1 mux act like a 2-input OR gate, if we connect D0 pin to B and D1 pin to A, with select connected to A.

Figure 5: Truth table of 2-input OR gate

Figure 6 below shows the implementation of 2-input OR gate using a 2:1 multilpexer:

Figure 6: Implementation of 2-input OR gate using 2:1 mux

Also read:

• NOT gate using 2:1 mux
• 2-input XOR gate using 2:1 mux
• 2-input NOR gate using 2:1 mux
• 2-input XNOR gate using 2:1 mux
• 2-input NAND gate using 2:1 mux
• 2-input AND gate using 2:1 mux

2-input XNOR gate using 2:1 mux

2-input XNOR gate using 2x1 mux: Figure 1 below shows the truth table of a 2-input XNOR gate. If we observe carefully, OUT equals B' when A is '0' and equals B when A is '1'. So, a 2-input XNOR gate can be implemented from a 2x1 mux, if we connect SEL pin to A, D0 to B' and D1 to B.

Figure 1: Truth table of 2-input XNOR gate

The implementation of 2-input XOR gate using a 2x1 mux is as shown in figure 2.

Figure 2: Implementation of 2-input XNOR gate using 2x1 mux

Similarly, we can observe the output for different values of B and follow the same steps to obtain the XNOR gate implementation.

8x1 multiplexer using 4x1 multiplexer

An 8x1 mux can be implemented using two 4x1 muxes and one 2x1 mux. 4 of the inputs can first be decoded using each 4-input mux using two least significant select lines (S0 and S1). The output of the two 4x1 muxes can be further multiplexed with the help of MSB of select lines at further stage. The implementation of 8x1 using 4x1 and 2x1 muxes is shown in figure 1 below:

2-input NOR gate using 2:1 mux

2-input NOR gate using 2x1 mux: Figure 1 below shows the truth table of a 2-input NOR gate. If we observe carefully, OUT equal B' when A is '0'. Similarly, OUT equals '0' when A is '1'. So, we can make a 2-input mux act like a 2-input NOR gate, if we connect SEL of mux to A, D0 to B' and D1 to '0'.

Figure 1: Truth table of 2-input NOR gate

Figure 2 shows the implementation of 2-input NOR gate using 2:1 mux.

Figure 2: Implementation of 2-input NOR gate using 2x1 mux

Similarly, we can connect B to select pin of mux and follow the same procedure of observation from truth table to get the NOR gate implemented.

Also read:

• NOT gate using 2:1 mux
• 2-input XOR gate using 2:1 mux
• 2-input XNOR gate using 2:1 mux
• 2-input OR gate using 2:1 mux
• 2-input NAND gate using 2:1 mux
• 2-input AND gate using 2:1 mux

Matchstick game bonanza!!

Problem: There are 'N' matchsticks placed on the table. You and your opponent are to pick any number of matchsticks between 1 and 5. Its your turn first. The one picking the last stick loses the game. You have to device a strategy such that you always win the game.Also, is there any starting number that cannot guarantee that you always win?

Solution: Here, you have to ensure that the control of the game always remains in your hands. Let us approach this problem from the last. The last matchstick has to be picked by your opponent so as to ensure your win. So, your last turn must ensure that there must be only one matchstick left on the table. If this is not the case, say there are 2 matchsticks. Then, your opponent will pick 1 stick and you are left with only one stick to pick and lose the game.

Similarly, one turn before last of your opponent, there must be more than 6 sticks so that he cannot leave you with 1 stick by picking 5 of them. But if there are more than 7 sticks, he may leave you with the same situation by leaving 7 on the table. If you leave 7, the other can at max pick 5 and min 1 leaving any number between 6 and 2 on the table. You can now pick the desired number leaving the last stick to be picked by him. Now, if there were 8 sticks on the table, your opponent would pick 1 leaving you with 7 sticks. :-( Similarly, one turn before, you should have left 13 sticks on the table.

So, your approach should be to leave (6M + 1) sticks on the table always. Ensure that at the end of each round, 6 less matchsticks are there on the table. (For instance, if your opponent picks 3 sticks, you also pick 3).

But there is a catch in this game, if there are already (6M + 1) sticks on the table initially, and its your turn first, you cannot ensure after your first turn the winning strategy. Now, the control goes into the hand of your opponent and you are at the verge of losing the game.

Similarly, 5 can be replaced with any number 'K'. At the end of each turn, you have to ensure there are (K+1)M + 1 sticks left on the table.