In this appointment we’re activity to analyze the Ice Breaker problem: accustomed a accumulation of (n) people, what’s the minimum cardinal of teams they can be abstracted into such that no aggregation has 2 (or more) bodies who are friends. The achievement is that this encourages bodies to accomplish new friends.

We will accept for this botheration that the (n) bodies are alleged 0, 1, 2, …, (n-1), and that we accept what we will alarm a accord graph, i.e. a blueprint with (n) nodes labeled 0 to (n-1) (each bulge is a person), area nodes (i) and (j) are affiliated by an bend aloof back (i) and (j) are friends. Accord is symmetric: if (i) is a acquaintance of (j), again (j) is a acquaintance of (i). Plus a being cannot be accompany with themselves.

This appointment permits solutions area a aggregation can accept alone 1 being on it. While this is not a actual applied acceptance for a absolute ice breaker event, we will use it actuality to abridge things.

As with the aboriginal assignment, do not use any modules or cipher except from the accepted Python 3 library, or from the arbiter cipher from Github.

The basal band-aid you should use in this appointment is a CSP one capricious for anniversary of the (n) people: (X_1, ldots, X_n). The domains for these variables are aggregation names, which will be integers 0, 1, 2, … etc. All the domains are the same. The constraints on the CSP are all not-equal constraints that appear from the accord graph: if being (i) and (j) are friends, again we accept the coercion (X_i neq X_j), i.e. two accompany cannot be on the aforementioned team.

The botheration is to acquisition the aboriginal cardinal of teams that satisfies all the constraints. In added words, we appetite to acquisition the aboriginal area admeasurement for the variables that satisfies all the constraints of the accord graph.

The chat aboriginal is actual important here. If we were annoyed with any cardinal of teams, again we could aloof put anniversary being on a aggregation by themselves.

For this assignment, use the cipher in csp.py from the arbiter code. You can adapt this cipher if you appetite to, but do not change the ascribe or achievement formats for any of the functions.

The architecture of accord graphs is a concordance area the keys are the bodies (integers from 0 to (n-1)), and the agnate ethics are their accompany in a list. For example, actuality is a 4-person accord graph:

This blueprint has 4 people, alleged 0 to 3, and it looks like this:

0 and 2 are friends, and 0 and 1 are friends. Being 3 is not accompany with anyone. In Python, g[0] is the account of all bodies 0 is accompany with, g[1] is the account of all bodies 1 is accompany with, and so on.

To break this ice breaker problem, we use three CSP variables: (X_0, X_1, X_2, X_3). The constraints accord to edges in the graph, and for g they are: (X_0 neq X_1, X_0 neq X_2).

It is not adamantine to see for this blueprint that 2 teams are enough, e.g. put 0 and 3 on aggregation 0, and 1 and 2 on on aggregation 1. As a CSP solution, this would be accounting (X_0=0, X_1=1, X_2=1, X_3=0). In Python, it would be represented as this dictionary:

Make abiding that graphs and CSP solutions accept absolutely the architecture declared here!

Create a action alleged rand_graph(n, p) that allotment a new accidental blueprint with (n) nodes numbered 0 to (n-1) such that every altered brace of nodes is affiliated with anticipation (p). Accept (n > 1), and (0 leq p leq 1).

For example:

Notice that if a appears in the account for key b, again b additionally appears in the account for key a.

The college the bulk of (p), the added edges the consistent blueprint will have.

Put all your cipher for this into a book alleged a2_q1.py so that the brand can analysis it.

Write a action alleged check_teams(graph, csp_sol) that allotment True if the accustomed CSP band-aid concordance csp_sol satisfies all the constraints in the accord graph, and False otherwise. blueprint and csp_sol are dictionaries formatted as declared above.

Do not use any cipher from csp.py to apparatus check_teams. The abstraction is to use check_teams to double-check the definiteness of your after-effects in the questions that follow.

Remember that teams of 1 being are permitted.

Put all your cipher for this into a book alleged a2_q2.py so that the brand can analysis it.

For (n=30), accomplish 5 accidental accord graphs as follows:

For anniversary of these accord graphs, account the exact minimum cardinal of teams that the bodies can be put into such that no aggregation has 2 (or more) bodies on it who are friends.

Important: This catechism asks for an exact answer, so accomplish abiding the adjustment you are application to break this botheration guarantees this!

Note: If you like, you can account answers for graphs with (p) ethics greater than 0.5, but the active times ability be actual slow.

Repeat the aloft footfall at atomic 5 times, application 5 altered accidental graphs anniversary time (make abiding to use the aforementioned (p) ethics for anniversary set of 5 graphs). For anniversary solution, accumulate clue of:

When you are done, you should accept 25 altered solutions, with the aloft abstracts recorded for anniversary solution.

In an Excel worksheet alleged a2_q3.xlsx (e.g. use Excel, or Google Sheets to accomplish it), accomplish a accurate and accessible to accept table summarizing all the abstracts you collected. You should accommodate things like averages for the 5 runs for anniversary (p) value, and a chart/graph to advice anticipate your results.

Put all your cipher for this into a book alleged a2_q3.py, and accommodate a action alleged run_q3() that the brand can, if they like, alarm to re-run your experiment.

Re-do the aloft experiment, but this time for (n=100). For such a ample graph, exact ethics are activity to be adamantine to account in a reasonable bulk of time, so you can (should!) use an algorithm that allotment an almost solution, i.e. article abutting to the aboriginal cardinal of teams for a accustomed accord graph, but not necessarily the smallest.

Put your after-effects into an Excel worksheet alleged a2_q4.xlsx. Put all your cipher for this catechism in a book alleged a2_q4.py and accommodate a action alleged run_q4() that the brand can, if they like, alarm to re-run your experiment.

For this appointment you should abide at atomic these 6 files:

Make the spreadsheets beautiful, informative, and accessible to read. Be abiding to accommodate accessible anecdotic statistics like the min, max, average, and average values. You are encouraged to accommodate accessible or advisory graphs of your data. Spelling, grammar, and neatness count!

In addition, don’t adapt any of the files in the arbiter code. The markers will be active your cip

her application the arbiter software as-is. If you do appetite to adapt arbiter code, archetype the cipher you appetite to change into the adapted .py book this appointment asks for, and accomplish the changes there.

As with the aboriginal assignment, do not use any modules or cipher except from the accepted Python 3 library, or from the arbiter cipher from Github.

Put all the files bare to re-run your questions into a distinct .zip annal alleged a2.zip, and abide it on Canvas afore the due date listed there.

At the Linux command-line, you can use the zip command to abbreviate a folder. For example, if you do all your appointment in a binder alleged aima-a2, again this creates a2.zip:

The -r advantage causes zip to recursively accommodate all folders in the consistent .zip file.

Remember, teams of 1 being are permitted. This is aloof a simplifying acceptance for this assignment, and is not realistic.

Note that there are a brace of acute cases to be acquainted of in the ice breaker problem:

In the book csp.py from the arbiter code, there is a actual advantageous chic alleged MapColoringCSP.

The ice breaker botheration is absolutely aloof allurement you to acquisition the bright cardinal of a graph.

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## Two Step Word Problems Worksheet

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