Monday, June 29, 2020
Saturday, June 27, 2020
Thursday, June 25, 2020
In the USA 93% of Afro-American homicide victims are killed by other Afro-Americans (link). However, I don't see BLM protesters protesting those victims, and I see nothing about it on the BLM website. So apparently their lives mattered little or none.
In The New York Times Michael D. Shear -- who previously worked 18 years for the Washington Compost -- wrote: "In the interview, Mr. Pence called Mr. Floyd’s death “a tragedy,” but his insistence that “all lives matter” is likely to be seen as a provocation by activists and others who say that phrase dilutes the issue and fails to acknowledge the specific threats that African-Americans still face at the hands of police officers in the United States" (link).
Pandering to BLM, both Shear and Taff failed to acknowledge the meaning of "all." "All" includes white, black, Asian, Hispanic, and every other possibility. Shear also ignores (1) the African-American homicide victims killed by other African-Americans, and (2) the specific threats that African-Americans still face at the hands of other African-Americans. When Shear believes it's fair to criticize Pence for what Pence doesn't say, he made it fair to criticize Shear for what he doesn't say.
Monday, June 22, 2020
The formula R/r + 1 is correct. However, the reference – author Y. Nishiyami – doesn't give the formula, much less prove it is true for any R and r. Also, Nishiyama says, “Separating revolution from rotation is helpful for understanding, but doing so does not provide a fundamental solution.” This latter part is not true. For a coin with radius r to make one revolution around a stationary coin with radius R, the center of the moving coin travels a circular path with radius R + r. (R + r)/r = R/r + 1. Likewise, the circumference of the path is 2*pi*(R + r)/(2*pi*r) = R/r + 1 times the circumference of the revolving coin. Rotation is irrelevant. [End]
The key to this solution is to focus on the center of the moving coin. The center travels a circular path with radius R + r and hence circumference 2*pi*(R+r) in order to make one full revolution around the stationary coin. How many rotations it makes, even zero, while doing so is irrelevant. (Rotation versus revolution.) If it "rolls without slipping or sliding," then the number of rotations will also be R/r + 1.
Saturday, June 20, 2020
Tuesday, June 16, 2020
Saturday, June 13, 2020
Dorn was shot to death, as much or more unjustly than George Floyd. Dorn was African-American. Didn’t his life matter, too? But there aren’t any protests for Dorn, no public kneeling in his memory.
Friday, June 5, 2020
Wednesday, June 3, 2020
His solution has recurring trios in the rows of every band and columns of every stack. For example, he entered 4 8 3 in rows 1-3 and 6 9 3 in columns 4-6. The order of digits within some trios aren't all the same. For example, there are 1 4 7 and 7 1 4 and 4 7 1.
The first digits I entered were as follows. I did the trio 2 6 1 first, then 4 8 5 and 7 3 9.
Then I filled the rest of the grid as follows, which shows many recurring identical trios.
I finished this full grid in only 3 minutes, 36 seconds. It satisfies the king's move and knight's move rules. However, it violates the other extra rule. Pairs of successive digits appear next to one another, e.g. 1 2 and 4 5. I tried tweaking to get rid of them, but gave up after 20 minutes.
Monday, June 1, 2020
- swapping two rows in a band
- swapping two columns in a stack
- rotating the entire grid 90, 180, or 270 degrees
- flipping the entire grid top to bottom or right to left
- transposing around the SW-NE diagonal or NW-SE diagonal.
The following is the second puzzle I showed on May 17, the one above after shuffling digits. It appears more random because 1 2 3 has been replaced by 6 4 7, and 3 4 5 by 7 2 8, and so forth. Yet there are still recurring trios in the rows of every band and columns of every stack. The digits are different, but the locations are the same.
I have since found ways to eliminate the recurring trios by rearranging a few numbers in the grid. The logic for doing so is rather complicated, but not trial & error. After such rearranging, the scrambler I described on May 17, or the transformations described above, could be used with the grid to create more puzzles with no recurring trios.
Part of the logic is shown as follows. Suppose the top band looks like the following. (It is the first stack above transposed.)
Every row in the left box is repeated in the other two. Swap the numbers 3, 6, 9 in rows 1 & 2. The result would be as follows, with the changes made in green. The recurring trios are eliminated without violating any Sudoku rules.
I suspect professional Sudoku puzzle makers use a different method involving trial & error. Example 1. Example 2. Whatever methods they do use are likely embodied in a computer program that could do all the trial & error very, very quickly. Do such programs screen for repeating trios? I suspect so.
By the way, every Sudoku solution I have ever seen has repeating pairs in 3x3 boxes aligned horizontally ("band") or vertically ("stack"). The following example is from here.
The pairs 1,7 and 2,8 repeat in rows 1-3; the pairs 1,4 and 8,9 in rows 4-6, the pairs 1,6 and 2,3 in rows 7-9. The pairs 9,1 and 2,4 repeat in columns 1-3; the pairs 5, 9 and 2,4 in columns 4-6, the pairs 1,8 and 2,7 in columns 7-9. Such repeating pairs are impossible to avoid.