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Put these numbers in order from least to greatest: 5, 7, 2, 3, 1, 4, 6, 8
The correct order is 1, 2, 3, 4, 5, 6, 7, 8 (you’re welcome) — for a possible score of 8 points. How many points would this response earn?
1, 2, 4, 5, 3, 7, 8, 6
and she then discusses various methods for scoring different answers.
What metrics might we devise to measure the discrepancy between an answer and the correct answer, and assign a grade – perhaps a % – to any given answer?
James Tanton (@jamestanton on Twitter) posted the following tweet February 5, 2018:
“Connect each of N red dots to each of N blue dots on a page with straight line segments. Let I(N) be the least number of intersection points that might be seen. I(1) = 0, I(2) = 0, I(3) = 1, I(4) = ?”
Compare this with the crossing number problem for the complete bipartite graph
Note that James Tanton has insisted on straight line segments joining blue and red dots, and not allowed more general curves.
We can certainly draw the complete bipartite graph with 4 red and 4 blue vertices with straight line segments for edges that cross in at most 4 places:
Since the crossing number for (straight line segments not required for edges) is 4 we cannot do better with straight lines, so I(4)=4.
What might be a solution to the calculation of I(N) for different N, and how might this relate to the crossing number for complete bipartite graphs?
More generally we can ask what is the minimum possible number of crossings when we join m red dots to each of n blue dots using only straight line segments; that is, the complete bipartite graph drawn with straight line segments for edges. Compare this with the known results on the crossing number of the complete bipartite graph (general curved edges allowed).
For example, has crossing number 2, and we can realize this with straight line edges:
Dr. James Tanton is a high profile mathematics educator, working in the U.S.
Originally form the city of Adelaide in Australia, he obtained his Ph.D. in Mathematics from Princeton University.
He is the author of Mathematics Galore! the prescribed text for MTH 487 Math Inquiry I, one of the cofounders of the Global Math Project, the instigator of Exploding Dots, a Mathematical Association of America Visiting Scholar, and an author of many books on the teaching and learning of mathematics. Together with students at St Mark’s School in Southborough, Massachusetts he holds the world record for folding paper: 10 miles of toilet paper folded 13 times.
The following message was created by a simple random substitution code:
“”ndyuji cjy yxndyuji kxnfxw vxkkcixk rkuji c lcfuxwo dh ndyuji vxwedyk, ncj ax c sdbxfhrt bco dh ujwfdyrnuji vuyytx cjy euie kneddt kwryxjwk wd ycwc ndttxnwudj cjy cjctokuk, jrvaxf wexdfo cjy ctixafc ctkd, ndyuji cjy yxndyuji cfx ujwfujkunctto ujwxfxkwuji wd kwryxjwk dh vdkw cixk, kd xjicixvxjw uk jdw rkrctto cj ukkrx hujyuji wuvx cjy kscnx uj wex nrffunrtrv vuiew ax vdfx dh cj ukkrx wexfx cfx kdvx lxfo iddy fxkdrfnxk cjy wxmwk wd rkx hdf ndyuji cjy yxndyuji kxnfxw vxkkcixk.””
That is, every letter of the alphabet was randomly replaced by a unique letter of the alphabet. This was not, therefore, a Caesar cipher.
This type of substitution code is susceptible to decoding via frequency counts.
See if you can decode the message.
To make it a little easier I have preserved the word structure.
Note: Zach observed that I had not, originally, coded the capitalized letters, and he cracked the coded message pretty quickly using that and his prior experience in code breaking.
Here’s another, similarly coded message for you to crack:
“gqj zug qjm krj uqkz unr rzd fex gqp ugfe jma euu yjn bad myj rmy and leu grs zqj jzm rgu qyr mya nqk jmk umr jzq jqe ean dle ugr rzd fex zqw uqa dmy jtd nqe erj fxu yjrj dlu qle ujd luc mys dnp myc dyj zua ndl eug buj rzd fex qes qbr kzq eeu ycu uwu yjz ugd rjq leu rjf xuy jrk euq neb gbt mnr jgu rrq cut qme uxm yjzu ruk dyx nur auk jlu kqf ruv qkz qnb sqr qle ujd knq kpm jmy nqa mxd nxu nzun umr qyd jzu ngu rrq cuj zun utd nut dnb dfj d qjj uga jjd knq kpf rmy cqt nui fuy kbq yqe brm r”
This time I made sure to put the message in lower case, removed punctuation, and then I split the coded message into blocks of 3 letters to destroy the word lengths.
- I produced this coded message using Mathematica®, and the code of John McLoone at Breaking Secret Codes with Mathematica®
- A free full cloud version of Mathematica®, with limited storage capability, is available at the Wolfram Development Platform (so schools and students do not need to purchase Mathematica® – they just need an internet connection).